Calculus Examples

$y = \arcsin (\arccos (t))$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \arcsin (t)$ and $g (t) = \arccos (t)$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $\arccos (t)$ .

$\frac{d}{d u} [\arcsin (u)] \frac{d}{d t} [\arccos (t)]$

Step 1.2

The derivative of $\arcsin (u)$ with respect to $u$ is $\frac{1}{\sqrt{1 - u^{2}}}$ .

$\frac{1}{\sqrt{1 - u^{2}}} \frac{d}{d t} [\arccos (t)]$

Step 1.3

Replace all occurrences of $u$ with $\arccos (t)$ .

$\frac{1}{\sqrt{1 - {\arccos (t)}^{2}}} \frac{d}{d t} [\arccos (t)]$

Step 2

The derivative of $\arccos (t)$ with respect to $t$ is $- \frac{1}{\sqrt{1 - t^{2}}}$ .

$\frac{1}{\sqrt{1 - {\arccos (t)}^{2}}} (- \frac{1}{\sqrt{1 - t^{2}}})$

Step 3

Simplify terms.

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Step 3.1

Multiply $\frac{1}{\sqrt{1 - {\arccos (t)}^{2}}}$ by $\frac{1}{\sqrt{1 - t^{2}}}$ .

$- \frac{1}{\sqrt{1 - {\arccos (t)}^{2}} \sqrt{1 - t^{2}}}$

Step 3.2

Combine using the product rule for radicals.

$- \frac{1}{\sqrt{(1 - {\arccos (t)}^{2}) (1 - t^{2})}}$

Step 3.3

Rewrite $1$ as $1^{2}$ .

$- \frac{1}{\sqrt{(1^{2} - {\arccos (t)}^{2}) (1 - t^{2})}}$

Step 4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \arccos (t)$ .

$- \frac{1}{\sqrt{(1 + \arccos (t)) (1 - \arccos (t)) (1 - t^{2})}}$

Step 5

Rewrite $1$ as $1^{2}$ .

$- \frac{1}{\sqrt{(1 + \arccos (t)) (1 - \arccos (t)) (1^{2} - t^{2})}}$

Step 6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = t$ .

$- \frac{1}{\sqrt{(1 + \arccos (t)) (1 - \arccos (t)) (1 + t) (1 - t)}}$